Foundations — Antiderivatives, the +C, and Standard Formulae

A function $F$ is an antiderivative of $f$ if $F'(x) = f(x)$. The indefinite integral $\int f(x)\,dx = F(x) + C$ denotes the whole family of antiderivatives, where $C$ is an arbitrary constant. Because $\dfrac{d}{dx}[F(x) + C] = f(x)$ for every constant $C$, the constant can never be recovered from $f$ alone — you need an extra condition (a boundary value) to fix it.

The integrals you must recall instantly:

• $\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ for $n \neq -1$
• $\int \dfrac{1}{x}\,dx = \log|x| + C$
• $\int e^x\,dx = e^x + C$, and $\int a^x\,dx = \dfrac{a^x}{\log a} + C$
• $\int \sin x\,dx = -\cos x + C$, $\int \cos x\,dx = \sin x + C$
• $\int \sec^2 x\,dx = \tan x + C$, $\int \csc^2 x\,dx = -\cot x + C$
• $\int \dfrac{dx}{1+x^2} = \tan^{-1}x + C$, $\int \dfrac{dx}{\sqrt{1-x^2}} = \sin^{-1}x + C$

If $\int f(x)\,dx = F(x) + C$, then for a linear inner argument $\int f(ax+b)\,dx = \dfrac{1}{a}\,F(ax+b) + C$. The $\dfrac{1}{a}$ is the leftover from $d(ax+b) = a\,dx$. This covers the whole table at once:

• $\int (ax+b)^n\,dx = \dfrac{(ax+b)^{n+1}}{a(n+1)} + C$
• $\int \dfrac{dx}{ax+b} = \dfrac{1}{a}\log|ax+b| + C$
• $\int e^{ax+b}\,dx = \dfrac{1}{a}e^{ax+b} + C$, $\int \sin(ax+b)\,dx = -\dfrac{1}{a}\cos(ax+b) + C$

Integration is linear: $\int [a\,f(x) + b\,g(x)]\,dx = a\int f(x)\,dx + b\int g(x)\,dx$. Pre-processing moves that pay off repeatedly:

• Improper fraction (degree of top $\geq$ bottom): do polynomial division first.
• Factorable numerator: cancel against the denominator.
• Split a single fraction into a sum of simpler ones.

Given $f'(x)$ and a single value $f(a) = k$: integrate to obtain $f(x) = F(x) + C$; substitute $x = a$ and set equal to $k$ to solve $C = k - F(a)$. The boundary condition removes the ambiguity of the constant.

Two recurring shapes: (1) composition — compute $(f\circ f)(x) = f(f(x))$ and simplify before integrating; (2) implicit definition — if $f(g(x)) = h(x)$, substitute $t = g(x)$, express $x$ in terms of $t$, and read off $f(t)$. Then integrate the resulting explicit function using the table.

Keep these collapses in reflex memory:

• **Half-angle of $1\pm\cos x$:** $1-\cos x = 2\sin^2\tfrac{x}{2}$, $1+\cos x = 2\cos^2\tfrac{x}{2}$; so $\dfrac{1-\cos x}{1+\cos x}=\tan^2\tfrac{x}{2}$, $\dfrac{1}{1-\cos x}=\tfrac12\csc^2\tfrac{x}{2}$, $\dfrac{1}{1+\cos x}=\tfrac12\sec^2\tfrac{x}{2}$.
• Power-reduction (double angle): $1-\cos 2x = 2\sin^2 x$, $1+\cos 2x = 2\cos^2 x$, $\sin 2x = 2\sin x\cos x$.
• Perfect square under a root: $1\pm\sin 2x=(\sin x\pm\cos x)^2$ and $1\pm\sin\theta=\left(\cos\tfrac\theta2\pm\sin\tfrac\theta2\right)^2$, so $\sqrt{1\pm\sin 2x}=|\sin x\pm\cos x|$ — keep the modulus; its sign depends on the interval.
• **$\sec\pm\tan$:** $\sec x+\tan x=\dfrac{1+\sin x}{\cos x}=\tan\!\left(\tfrac\pi4+\tfrac x2\right)$, $\sec x-\tan x=\tan\!\left(\tfrac\pi4-\tfrac x2\right)$.
• Harmonic form: $a\sin x+b\cos x=\sqrt{a^2+b^2}\,\sin(x+\alpha)$, so its extreme values are $\pm\sqrt{a^2+b^2}$.
• **Weierstrass $t=\tan\tfrac{x}{2}$:** $\sin x=\dfrac{2t}{1+t^2}$, $\cos x=\dfrac{1-t^2}{1+t^2}$, $dx=\dfrac{2\,dt}{1+t^2}$ — turns any rational function of $\sin x,\cos x$ into a rational function of $t$.